作業資料不提交到數據庫表

Question

我有一個論壇，用戶可以輸入他們正在尋找的工作，這將提交到數據庫，然後顯示在下一頁。只有我無法獲取任何數據才能上傳，我不知道爲什麼。

我也在努力尋找錯誤檢查的方法。有任何想法嗎？

// Check for job submission 
     if(isset($_POST['submit'])) 

      //empty error array 
      $error = array(); 

     // check for a things in the field 

     if(empty($_POST['job'])) 
     { 
      $error[] = 'Please fill in all required fields'; 

     } 

     // iff there are no errors, insert the job listing into the database. 
     // otherwies, display error. 
     if(sizeof($error) == 0) 
     { 
      // insert job listing 
      $query = "INSERT INTO job (
           job_id, 
           user_id, 
           jobtitle, 
           company, 
           location, 
           summary, 
           responsibilities, 
           skills 
           ) VALUES (
            null, 
            '{$_SESSION['user_id']}', 
            '{$_POST['jobtitle']}', 
            '{$_POST['company']}', 
            '{$_POST['location']}', 
            '{$_POST['summary']}', 
            '{$_POST['responsibilities']}', 
            '{$_POST['skills']}', 
            NOW() 
            )"; 

      $result = mysqli_query($dbc, $query) or die('Query failed: ' . mysqli_error($dbc)); 

      // display a confirmation 
      echo "<div class=\"alert alert success\">Your job listing has been added</div>"; 

     } else { 

      // display error message 
      foreach($error as $value) 
      { 
       echo "<div class=\"alert alert-danger\"{$value}</div>"; 
      } 
     } 

     ?> 

     <!-- Job listing Form --> 
     <form method="post" action="listings.php"> 
      <div class="form-group"> 
       <label> Job Title </label> 
       <input name ="jobtitle" type="text" class="jobform"/> 

       <label>Company/Organization</label> 
       <input name="company" type="text" class="jobform"/> 

       <label> Job Location </label> 
       <input name ="location" type="text" class="jobform"/> 

       <label> Job Responsibilities </label> 
       <textarea name="summary" rows="8" cols="20" class="jobfourm"></textarea> 

       <label> Job Responsibilities </label> 
       <textarea name="responsibilities" rows="8" cols="20" class="jobfourm"></textarea> 

       <label> Job Skills </label> 
       <textarea name="skills" rows="8" cols="20" class="jobforum"></textarea> 

      </div> 

     <div class="form-group"> 
        <input name="submit" type="submit" value="Submit" class="btn btn-large btn-primary" /> 
       </div> 

     </form> 

    </div> 

Answer 1

我的賭注是你的查詢：

(
           job_id, 
           user_id, 
           jobtitle, 
           company, 
           location, 
           summary, 
           responsibilities, 
           skills 
           ) VALUES (
            null, 
            '{$_SESSION['user_id']}', 
            '{$_POST['jobtitle']}', 
            '{$_POST['company']}', 
            '{$_POST['location']}', 
            '{$_POST['summary']}', 
            '{$_POST['responsibilities']}', 
            '{$_POST['skills']}', 
            NOW() 

什麼應該JOB_ID，你傳遞無效。現在，我假設所有的工作都必須有工作編號，對嗎？你需要實際傳遞一個有效的ID，因爲我打賭錢（或代表），這是表中的一個不可空字段。另外，您已經在您的列名稱參數中未聲明的值中添加了一列。